States of Matter — Kinetic Theory, Gas Laws for Class 11

What Are States of Matter, Really?

Matter exists in three common states — solid, liquid, and gas — but what determines which state a substance prefers? The answer is a constant tug-of-war between two forces: intermolecular forces (pulling molecules together) and thermal energy (pushing them apart).

When thermal energy wins, molecules spread out — you get a gas. When intermolecular forces win, molecules stay packed — you get a solid. Liquids are the contested middle ground.

This chapter in NCERT Class 11 (Chapter 5) is deceptively important. The gas laws, kinetic theory, and van der Waals equation form the foundation of physical chemistry. JEE Main has pulled questions from this chapter in nearly every sitting since 2019.

Key Terms and Definitions

Intermolecular forces — attractive forces between molecules. These include London dispersion forces, dipole-dipole interactions, and hydrogen bonds. Stronger forces → higher boiling point, more likely to be liquid or solid at room temperature.

Thermal energy — energy associated with random motion of molecules. Proportional to absolute temperature (in Kelvin, always).

Pressure (P) — force per unit area exerted by gas molecules colliding with container walls. SI unit: Pascal (Pa). We also use atm, bar, mm Hg (torr).

1 atm = 101.325 kPa = 760 mm Hg = 1.01325 bar

Volume (V) — space occupied by gas. SI unit: m³, but we commonly use litres (L) or mL.

Temperature (T) — always use Kelvin in gas law calculations. T(K) = T(°C) + 273.15. This is the single most common source of errors in exams.

Molar mass (M) — mass of one mole of substance in g/mol.

Compressibility factor (Z) — ratio PV/nRT. For an ideal gas, Z = 1. Real gases deviate from this.

The Gas Laws — One at a Time

Boyle’s Law (1662)

At constant temperature and amount, pressure and volume are inversely proportional.

PV = \text{constant}

P_1V_1 = P_2V_2 \quad \text{(at constant T, n)}

Why does this work? When we compress a gas, the same number of molecules now hit the walls of a smaller container more frequently. More collisions per second → higher pressure.

The graph of P vs 1/V gives a straight line through the origin. The P vs V graph is a hyperbola (rectangular hyperbola, to be precise in JEE terms).

Charles’ Law (1787)

At constant pressure, volume is directly proportional to absolute temperature.

\frac{V}{T} = \text{constant}

\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad \text{(at constant P, n)}

Extrapolating the V vs T graph to V = 0 gives T = −273.15°C = 0 K. This is how absolute zero was conceptually established — the temperature at which an ideal gas would occupy zero volume.

NCERT defines 0°C as 273.15 K, but in problems they often use 273 K for simpler numbers. JEE questions specify — read carefully.

Gay-Lussac’s Law (Pressure-Temperature Law)

At constant volume, pressure is directly proportional to absolute temperature.

\frac{P}{T} = \text{constant}

\frac{P_1}{T_1} = \frac{P_2}{T_2} \quad \text{(at constant V, n)}

Avogadro’s Law

At constant temperature and pressure, equal volumes of all gases contain equal numbers of molecules.

V \propto n \quad \text{(at constant T, P)}

\frac{V_1}{n_1} = \frac{V_2}{n_2}

At STP (0°C, 1 bar — the new IUPAC standard), one mole of ideal gas occupies 22.7 L. At the old STP (0°C, 1 atm), it was 22.4 L. NCERT Class 11 uses 22.4 L — stick with that for board exams, and check what the JEE question states.

The Ideal Gas Equation

Combining Boyle’s, Charles’, and Avogadro’s laws:

PV = nRT

Where $R = 8.314 \text{ J mol}^{-1}\text{K}^{-1} = 0.0821 \text{ L atm mol}^{-1}\text{K}^{-1}$

Which value of R to use? If P is in atm and V in litres → use 0.0821. If working in SI units (P in Pa, V in m³) → use 8.314. Mixing units is the fastest way to get a wrong answer.

Kinetic Theory of Gases

The kinetic molecular theory explains why ideal gas laws work by making five assumptions:

Gas consists of large number of molecules in continuous random motion.
Volume of individual molecules is negligible compared to container volume.
No attractive or repulsive forces between molecules (this is the “ideal” part).
Collisions are perfectly elastic — no kinetic energy is lost.
Average kinetic energy of molecules is proportional to absolute temperature.

From these assumptions, we derive the kinetic gas equation:

PV = \frac{1}{3}mN\overline{u^2}

Where $m$ = mass of one molecule, $N$ = number of molecules, $\overline{u^2}$ = mean square speed

Molecular Speeds

Not all molecules move at the same speed. The Maxwell-Boltzmann distribution describes the spread of speeds.

u_{rms} = \sqrt{\frac{3RT}{M}} \qquad \text{(root mean square speed)}

u_{mp} = \sqrt{\frac{2RT}{M}} \qquad \text{(most probable speed)}

\overline{u} = \sqrt{\frac{8RT}{\pi M}} \qquad \text{(average speed)}

The ratio $u_{mp} : \overline{u} : u_{rms} = 1 : 1.128 : 1.225$

JEE Main frequently asks you to rank these three speeds or calculate their ratio. Remember: most probable < average < rms. A helpful mnemonic: PAR (Probable < Average < Root mean square).

Kinetic Energy and Temperature

Average kinetic energy per molecule = $\frac{3}{2}k_BT$ , where $k_B$ = Boltzmann constant.

Average kinetic energy per mole = $\frac{3}{2}RT$

This is why temperature (in Kelvin) is literally a measure of average kinetic energy. At absolute zero, all molecular motion (theoretically) stops.

Real Gases and van der Waals Equation

Ideal gas assumes molecules have no volume and no intermolecular forces. Real molecules do both. At high pressure and low temperature, real gases deviate significantly from ideal behavior.

Why does real gas deviate?

At high pressure, molecules are close together — their own volume matters (can’t compress indefinitely).
At low temperature, molecules move slowly — intermolecular attractions become significant.

\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT

$a$ = measure of intermolecular attractions (pressure correction)
$b$ = excluded volume per mole (volume correction, ≈ 4 × actual molecular volume)

Correction logic:

We add $\frac{an^2}{V^2}$ to P because real pressure is less than ideal (intermolecular attractions reduce wall-hitting force).
We subtract $nb$ from V because the free volume available to molecules is less than the container volume.

Compressibility Factor (Z)

Z = \frac{PV}{nRT} = \frac{PV_m}{RT}

Z = 1 → ideal gas behavior
Z < 1 → gas is more compressible than ideal (attractive forces dominate — common at moderate pressure)
Z > 1 → gas is less compressible than ideal (repulsive forces/volume effects dominate — high pressure)

H₂ and He show Z > 1 at almost all pressures because their $a$ values are very small (weak intermolecular forces). Most other gases show Z < 1 first, then Z > 1 at very high pressure.

Solved Examples

Example 1 — Easy (CBSE Level)

A gas occupies 500 mL at 27°C and 1 atm. What volume does it occupy at 127°C and 2 atm?

Using the combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

Convert temperatures: T₁ = 300 K, T₂ = 400 K

V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{1 \times 500 \times 400}{300 \times 2} = \frac{200000}{600} = 333.3 \text{ mL}

Example 2 — Medium (JEE Main Level)

Calculate the rms speed of nitrogen molecules at 27°C. (M = 28 g/mol, R = 8.314 J/mol·K)

u_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}}

= \sqrt{\frac{7482.6}{0.028}} = \sqrt{267235.7} \approx 517 \text{ m/s}

Note: M must be in kg/mol (0.028), and R in J/mol·K to get speed in m/s.

Using M = 28 (in g/mol) without converting to kg/mol gives a speed 1000× too large. Always check your units when calculating molecular speeds.

Example 3 — Hard (JEE Advanced Level)

For a van der Waals gas, at what temperature will it behave most like an ideal gas: very low T, moderate T, or very high T?

At very high T, thermal energy dominates over intermolecular forces (large $a$ term becomes negligible), and the free volume available is large compared to $nb$ correction. Z → 1. So the answer is very high temperature.

This is the Boyle temperature concept: $T_B = \frac{a}{Rb}$ . At this temperature, the attractive and repulsive deviations cancel, giving near-ideal behavior over a range of pressures.

Exam-Specific Tips

CBSE Board Pattern: Chapter 5 carries questions worth 5-7 marks in most board papers. Numericals on combined gas law, ideal gas equation, and molar mass determination from density are standard. Definition-based questions (state each gas law with conditions) appear in 2-mark segments. Derivation of $PV = nRT$ from Boyle’s and Charles’ laws is a common 3-mark question.

JEE Main Weightage: This chapter averages 1-2 questions per paper. High-frequency topics: compressibility factor (Z vs P graphs), kinetic energy calculations, rms/average/most probable speed ratios, van der Waals corrections. JEE Main 2024 January Session had a question on comparing Z values of H₂ and CO₂ at the same conditions.

For density-molar mass problems: The ideal gas equation rearranges to:

M = \frac{dRT}{P}

where $d$ is density in g/L and P in atm. This form appears constantly in board practicals and JEE numericals.

Common Mistakes to Avoid

Mistake 1: Using Celsius instead of Kelvin. Charles’ law and the ideal gas equation require absolute temperature. T = 0°C is NOT zero temperature — it’s 273 K. If you use 0 in the denominator, your calculation collapses. Always convert first.

Mistake 2: Mixing pressure and volume units with the wrong R. R = 8.314 J/mol·K only when P is in Pa and V in m³. For P in atm and V in L, use R = 0.0821 L·atm/mol·K. Using 8.314 with atm and litres gives a nonsense answer.

Mistake 3: Forgetting the pressure correction sign in van der Waals. The equation adds $\frac{an^2}{V^2}$ to P. Students sometimes subtract it. Remember: real gas exerts less pressure than ideal because attractive forces pull molecules back from the wall — so we must add a term to get the ideal P.

Mistake 4: Confusing $u_{rms}$ with average speed. $u_{rms} = \sqrt{\frac{3RT}{M}}$ , not $\sqrt{\frac{2RT}{M}}$ (that’s $u_{mp}$ ). The factor under the root is 3 for rms, 2 for most probable, and $\frac{8}{\pi}$ for average. These are frequently swapped under exam pressure.

Mistake 5: Applying Avogadro’s 22.4 L without checking STP definition. Old STP: 0°C, 1 atm → 22.4 L/mol. New IUPAC STP: 0°C, 1 bar → 22.7 L/mol. NCERT Class 11 still uses 22.4 L in most examples. But if a JEE question specifies 1 bar, use 22.7 L.

Practice Questions

Q1. A balloon is filled with 2 L of helium at 27°C and 1 atm. It rises to an altitude where pressure is 0.5 atm and temperature is −23°C. What is the new volume?

Using combined gas law: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

T₁ = 300 K, T₂ = 250 K, P₁ = 1 atm, P₂ = 0.5 atm, V₁ = 2 L

$V_2 = \frac{1 \times 2 \times 250}{300 \times 0.5} = \frac{500}{150} = 3.33 \text{ L}$

Q2. Calculate the molar mass of a gas if 1.5 g of it occupies 800 mL at 27°C and 1 atm.

PV = nRT → n = PV/RT = (1 × 0.8) / (0.0821 × 300) = 0.8/24.63 = 0.0325 mol

M = mass/moles = 1.5/0.0325 = 46.2 g/mol (approximately NO₂ or C₂H₅OH)

Q3. At what temperature will the rms speed of oxygen molecules equal the rms speed of hydrogen molecules at 300 K?

Set $u_{rms}$ equal: $\sqrt{\frac{3RT_1}{M_{O_2}}} = \sqrt{\frac{3R \times 300}{M_{H_2}}}$

$\frac{T_1}{32} = \frac{300}{2}$

$T_1 = 150 \times 32 = 4800 \text{ K}$

Q4. For a van der Waals gas with a = 3.6 L²·atm/mol² and b = 0.04 L/mol, calculate the pressure of 2 mol of gas in a 10 L container at 300 K.

$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$

Pressure correction: $\frac{3.6 \times 4}{100} = 0.144$ atm

Volume correction: $V - nb = 10 - (2 \times 0.04) = 9.92$ L

$\left(P + 0.144\right)(9.92) = 2 \times 0.0821 \times 300 = 49.26$

$P + 0.144 = 4.966$

$P = 4.822 \text{ atm}$

Compare to ideal: P = nRT/V = 49.26/10 = 4.926 atm. Real gas gives lower pressure (attractive forces).

Q5. The ratio of average kinetic energy of O₂ to N₂ at the same temperature is: (a) 7:8 (b) 1:1 (c) 32:28 (d) 28:32

Answer: (b) 1:1

Average KE per mole = 3RT/2. This depends only on temperature, not on the identity or molar mass of the gas. At the same temperature, all ideal gases have the same molar kinetic energy.

Q6. At high pressure, a real gas has Z > 1. What does this mean physically?

Z > 1 means PV > nRT, i.e., the gas is harder to compress than an ideal gas. At high pressure, molecules are so close together that repulsive forces and the finite volume of molecules dominate. The gas resists further compression more than an ideal gas would.

Q7. A container holds a mixture of 16 g O₂ and 4 g He at a total pressure of 5 atm. Find the partial pressure of each gas.

Moles of O₂ = 16/32 = 0.5 mol; Moles of He = 4/4 = 1 mol; Total = 1.5 mol

Mole fraction of O₂ = 0.5/1.5 = 1/3; Mole fraction of He = 1/1.5 = 2/3

$P_{O_2} = \frac{1}{3} \times 5 = 1.67 \text{ atm}$ ; $P_{He} = \frac{2}{3} \times 5 = 3.33 \text{ atm}$

Q8. Why is the most probable speed less than the rms speed?

The Maxwell-Boltzmann distribution is not symmetric — it has a tail at high speeds. The mean square (and hence rms) is pulled upward by the small fraction of very fast molecules. The most probable speed corresponds to the peak of the distribution, which is at a lower value. Mathematically: $u_{mp} = \sqrt{2RT/M}$ vs $u_{rms} = \sqrt{3RT/M}$ , and $\sqrt{2} < \sqrt{3}$ .

FAQs

Why do gases mix completely but liquids often don’t? Gas molecules have negligible intermolecular forces compared to thermal energy — they distribute randomly throughout any available volume. Liquids have significant intermolecular forces, so “like dissolves like”: polar liquids mix with polar, non-polar with non-polar.

What is the significance of the van der Waals constant ‘a’? The constant $a$ measures the strength of intermolecular attractions. A higher $a$ means stronger attractions, a higher boiling point, and greater deviation from ideal behavior at low temperatures. NH₃ has a high $a$ (4.17) due to hydrogen bonding; He has a = 0.034 because it has almost no intermolecular forces.

Why is absolute zero (0 K) considered unattainable? As we cool a gas, we remove thermal energy. But at 0 K, there would be zero thermal energy — molecules would have no kinetic energy at all. Quantum mechanics tells us this violates the uncertainty principle. Experimentally, you can get extremely close to 0 K but never actually reach it.

What happens to Z at very high pressure and why? At very high pressure, the volume correction ( $nb$ ) dominates. Molecules are so crowded that excluded volume becomes significant — the gas cannot be compressed as easily as ideal. Z rises above 1. For gases like H₂ and He, this happens at even moderate pressures because their attractive forces are negligibly weak.

How is Dalton’s Law of Partial Pressures related to kinetic theory? Kinetic theory assumes no intermolecular forces between molecules. So molecules of different gases in a mixture don’t “know” about each other — each behaves as if it alone occupies the container. Each gas exerts pressure independently. Total pressure is simply the sum of all partial pressures.

Why does the Maxwell-Boltzmann distribution shift and flatten at higher temperatures? At higher T, more molecules have higher kinetic energy — the peak shifts right (higher most probable speed) and the curve flattens and broadens. The area under the curve stays constant (total number of molecules doesn’t change), so higher peak temperature means lower peak height.

Is the ideal gas equation exact for any real gas? No real gas is perfectly ideal, but noble gases like He and Ne come closest at room temperature and low pressure. At low pressure (molecules far apart) and high temperature (thermal energy dominates attractions), most gases approximate ideal behavior well. The ideal gas equation is a limiting law — real gases approach it as $P \to 0$ .